The Stacks project

Lemma 9.22.3. Let $L/K$ be a Galois extension with Galois group $G$. Let $\Lambda $ be the set of finite Galois subextensions, i.e., $\lambda \in \Lambda $ corresponds to $L/L_\lambda /K$ with $L_\lambda /K$ finite Galois with Galois group $G_\lambda $. Define a partial ordering on $\Lambda $ by the rule $\lambda \geq \lambda '$ if and only if $L_\lambda \supset L_{\lambda '}$. Then

  1. $\Lambda $ is a directed partially ordered set,

  2. $L_\lambda $ is a system of $K$-extensions over $\Lambda $ and $L = \mathop{\mathrm{colim}}\nolimits L_\lambda $,

  3. $G_\lambda $ is an inverse system of finite groups over $\Lambda $, the transition maps are surjective, and

    \[ G = \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } G_\lambda \]

    as a profinite group, and

  4. each of the projections $G \to G_\lambda $ is continuous and surjective.

Proof. Every subfield of $L$ containing $K$ is separable over $K$ (follows immediately from the definition). Let $S \subset L$ be a finite subset. Then $K(S)/K$ is finite and there exists a tower $L/E/K(S)/K$ such that $E/K$ is finite Galois, see Lemma 9.16.5. Hence $E = L_\lambda $ for some $\lambda \in \Lambda $. This certainly implies the set $\Lambda $ is not empty. Also, given $\lambda _1, \lambda _2 \in \Lambda $ we can write $L_{\lambda _ i} = K(S_ i)$ for finite sets $S_1, S_2 \subset L$ (Lemma 9.7.5). Then there exists a $\lambda \in \Lambda $ such that $K(S_1 \cup S_2) \subset L_\lambda $. Hence $\lambda \geq \lambda _1, \lambda _2$ and $\Lambda $ is directed (Categories, Definition 4.21.4). Finally, since every element in $L$ is contained in $L_\lambda $ for some $\lambda \in \Lambda $, it follows from the description of filtered colimits in Categories, Section 4.19 that $\mathop{\mathrm{colim}}\nolimits L_\lambda = L$.

If $\lambda \geq \lambda '$ in $\Lambda $, then we obtain a canonical surjective map $G_\lambda \to G_{\lambda '}$, $\sigma \mapsto \sigma |_{L_{\lambda '}}$ by Lemma 9.21.8. Thus we get an inverse system of finite groups with surjective transition maps.

Recall that $G = \text{Aut}(L/K)$. By Lemma 9.22.2 the restriction $\sigma |_{L_\lambda }$ of a $\sigma \in G$ to $L_\lambda $ is an element of $G_\lambda $. Moreover, this procedure gives a continuous surjection $G \to G_\lambda $. Since the transition mappings in the inverse system of $G_\lambda $ are given by restriction also, it is clear that we obtain a canonical continuous map

\[ G \longrightarrow \mathop{\mathrm{lim}}\nolimits _{\lambda \in \Lambda } G_\lambda \]

Continuity by definition of limits in the category of topological groups; recall that these limits commute with the forgetful functor to the categories of sets and topological spaces by Topology, Lemma 5.30.3. On the other hand, since $L = \mathop{\mathrm{colim}}\nolimits L_\lambda $ it is clear that any element of the inverse limit (viewed as a set) defines an automorphism of $L$. Thus the map is bijective. Since the topology on both sides is profinite, and since a bijective continuous map of profinite spaces is a homeomorphism (Topology, Lemma 5.17.8), the proof is complete. $\square$

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